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ISSN 1472-2739 (on-line) 1472-2747 (printed)
Algebraic & Geometric Topology
Volume 2 (2002) 1051–1060
Published: 5 November 2002
ATG
The slicing number of a knot
Charles Livingston
Abstract An open question asks if every knot of 4–genus gs can be
changed into a slice knot by gs crossing changes. A counterexample is
given.
AMS Classification 57M25; 57N70
Keywords Slice genus, unknotting number
A question of Askitas, appearing in [O, Problem 12.1], asks the following: Can
a knot of 4–genus gs always be sliced (made into a slice knot) by gs crossing
changes? If we let us (K) denote the slicing number of K , that is, the minimum
number of crossing changes that are needed to convert K into a slice knot, one
readily shows that gs (K) ≤ us (K) for all knots, with equality if gs (K) = 0.
Hence, the problem can be restated as asking if gs (K) = us (K) for all K .
We will show that the knot 74 provides a counterexample; gs (74 ) = 1 but
no crossing change results in a slice knot: us (74 ) = 2. It is interesting to
note that 74 already stands out as an important example. The proof that its
unknotting number is 2, not 1, resisted early attempts [N]; ultimately, Lickorish
[L] succeeded in proving that it cannot be unknotted with a single crossing
change.
As noted by Stoimenow in [O], if one attempts to unknot a knot of 4–ball genus
gs instead of converting it into a slice knot, more than gs crossing changes may
be required. This is obviously the case with slice knots. For a more general
example, let T denote the trefoil knot. One has gs (n(T # − T )#mT ) = m, but
u(n(T # − T )#mT ) = m + 2n, where u denotes the unknotting number.
The unknotting number, though itself mysterious, appears much simpler than
the slicing number. Many of the three–dimensional tools that are available
for studying the unknotting number do not apply to the study of the slicing
number. As we will see, even for this low crossing knot, 74 , the computation
of its slicing number is far more complicated than its unknotting number.
In the last section of this paper we introduce a new slicing invariant, Us (K),
that takes into account the sign of crossing changes used to convert a knot K
c Geometry & Topology Publications
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Charles Livingston
into a slice knot. This invariant is more closely related to the 4–genus and
satisfies
gs (K) ≤ Us (K) ≤ us (K).
It seems likely that there are knots K for which gs (K) 6= Us (K), and 74 seems
a good candidate, but we have been unable to verify this.
A good reference for the knot theory used here, especially surgery descriptions
of knots, crossing changes and branched coverings, is [R]. A reference for 4–
dimensional aspects of knotting and also for the linking form of 3–manifolds is
[G]. A careful analysis of the interplay between crossing changes and the linking
form of the 2–fold branched cover of a knot appears in [L], which our work here
generalizes. Different aspects of the relationship between crossing changes and
4–dimensional aspects of knotting appear in [CL]. A general discussion of slicing
operations is contained in [A].
1
Background
Our goal is to prove that a single crossing change cannot change 74 into a slice
knot. The key results concerning slice knots that we will be using are contained
in the following theorem; details of the proof can be found in [CG, G, R].
Theorem 1.1 If K is slice then:
(1) ∆K (t) = ±f (t)f (t−1 ) for some polynomial f , where ∆K (t) is the Alexander polynomial.
(2) |H1 (M (K), Z)| = n2 for some odd n, where M (K) is the 2–fold branched
cover of S 3 branched over K .
(3) There is a subgroup H ⊂ H1 (M (K), Z) such that |H|2 = |H1 (M (K), Z)|
and the Q/Z–valued linking form β defined on H1 (M (K), Z) vanishes
on H .
Our analysis of 74 will focus on the 2–fold branched cover, M (74 ), and its
linking form. This is much as in Lickorish’s unknotting number argument.
However, in our case the necessary analysis of the 2–fold branched cover can
only be achieved by a close examination of the infinite cyclic cover. In the next
two sections we examine the 2–fold branched cover; in Section 4 we consider
the infinite cyclic cover.
Algebraic & Geometric Topology, Volume 2 (2002)
The slicing number of a knot
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Crossing Changes and Surgery
If a knot K 0 is obtained from K by changing a crossing, surgery theory as
described in [R] quickly gives that the 2-fold cover of K 0 , M (K 0 ), can be
obtained from M (K) by performing integral surgery on a pair curves, say S1
and S2 , in S 3 . It is also known [L, Mo] that M (K 0 ) can be obtained from
M (K) by performing p/2 surgery on a single curve, say T , in S 3 . Here it will
be useful to observe that T can be taken to be S1 , as we next describe.
A crossing change is formally achieved as follows. Let D be a disk meeting
K transversely in two points. A neighborhood of D is homeomorphic to a
3–ball, B , meeting K in two trivial arcs. In one view, a crossing change is
accomplished by performing ±1 surgery on the boundary of D, say S . Then
S lifts to give the curves S1 and S2 in M (K). In the other view, the crossing
change is accomplished by removing B from S 3 and sewing it back in with
one full twist. The 2–fold branched cover of B is a single solid torus, a regular
neighborhood of its core T . A close examination shows the surgery coefficient
in this case is p/2 for some odd p.
The lift of D to the 2-fold branched cover is an annulus with boundary the
union of S1 and S2 and core T (the lift of an arc, τ , on D with endpoints the
two points of intersection of D with K ). Clearly T is isotopic to either Si , as
desired.
The following generalization of these observations will be useful. Rather than
put a single full twist between the strands when replacing B , n full twists can
be added. This is achieved by performing ±1/n surgery on S and hence the
2–fold branched cover is obtained by performing p/n surgery on the Si for some
p, or, by a similar analysis, by performing p0 /2n surgery on T for some p0 .
3
Results based on the 2–fold branched cover of 74
On the left in Figure 1 the knot 74 is illustrated. Basic facts about 74 include
that it has 3–sphere genus 1 and that its Alexander polynomial is ∆74 = 4t2 −
7t + 4. Since the Alexander polynomial is irreducible, 74 is not slice, so we have
gs (74 ) = 1. Also, 74 is the 2–bridge B(4, −4), and hence from the continued
fraction expansion it has 2–fold branched cover the lens space, L(15, 4).
The right diagram in Figure 1 represents a surgery diagram of 74 . According
to [R], surgery on the link K q K 0 with coefficient −1 and −2 yields S 3 . Also
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Charles Livingston
U
τ
74
K'
-2
K
-1
τ
Figure 1: The knot 74
according to [R] the component K 0 could be ignored in the diagram if −1/2
surgery is performed on K instead. In both cases the effect is to put two full
right handed twists in the two strands passing through K .
Notice that U is unknotted. After surgery is performed, U is converted into
the knot 74 .
If a knot J is obtained from 74 by a single crossing change, that change is
achieved via a disk D meeting 74 in two points, marked schematically by the
two dots in the right hand diagram. The path on D joining those two points
is denoted τ , a portion of which is also indicated schematically. By sliding τ
over K repeatedly it can be arranged that τ misses the small disk bounded by
K 0 meeting K in one point. The boundary of D will be denoted S and one of
its lift to the 2–fold branched cover of S 3 over U (this cover is again S 3 since
U is unknotted) will be denoted S1 . Neither D nor S is drawn in the figure.
Since two full twists on the unknot U convert it into 74 , the 2–fold branched
cover of S 3 branched over 74 is, by our earlier discussion, obtained from S 3 by
surgery on a single lift of K , say K1 , with surgery coefficient of the form p/4
for some p. Since we know that the cover is L(15, 4), we actually know that
p = 15, though for the argument that follows, simply knowing that p = ±15
would be sufficient.
Theorem 3.1 If the linking number of K1 and S1 in S 3 is divisible by 15
then J is not slice.
Proof Suppose that the linking number is divisible by 15. Since 15/4 surgery
is performed on K1 , after repeatedly sliding S1 over K1 it can be arranged
that the linking number of K1 and S1 is 0. The 2–fold cover of S 3 branched
Algebraic & Geometric Topology, Volume 2 (2002)
The slicing number of a knot
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over J , that is M (J), is obtained from S 3 by performing 15/4 surgery on K1
and p/2 surgery on S1 for some odd p.
If J is slice, the order of the homology of M (J) is an odd square and hence
p = ±52k+1 32j+1 q 2 , where q is relatively prime to 30.
We have that H1 (M (J), Z) = Z15 ⊕ Z|p| generated by the meridians of K1 and
S1 , denoted m1 and m2 , respectively.
The Q/Z–valued linking form, β , on H1 (M (J), Z) is orthogonal with respect to
this direct sum decomposition since the linking number is now 0. Furthermore,
from the surgery description we have that β(m1 , m1 ) = 4/15 and β(m2 , m2 ) =
2/p. The 5–torsion in H1 (M (J), Z) is isomorphic to Z5 ⊕ Z52k+1 , generated by
n1 = 3m1 and n2 = 32j+1 q 2 m2 . A quick calculation shows that β(n1 , n1 ) = 2/5
and β(m2 , m2 ) = 2(32j+1 q 2 )2 /p = ±2(32j+1 q 2 )/52k+1 .
If J is slice, the linking form on the 5–torsion vanishes on a subgroup of order
5k+1 . Suppose that n1 + x5l n2 has self–linking 0 ∈ Q/Z, where x is relatively
prime to 5. Then we would have
2 2x2 52l (32j+1 q 2 )
= 0 ∈ Q/Z.
±
5
52k+1
This implies that l = k , and hence that 2±2x2 32j+1 q 2 ≡ 0 mod 5. Letting q 0 =
x3j q , this can be rewritten as 2 ± 2(3q 0 2 ) ≡ 0 mod 5, or that 2 ≡ ∓q 0 2 mod 5.
However, the only squares modulo 5 are ±1, so this is impossible.
It follows from this that any element of self–linking 0 must be of the form x5l n2
for some l and x relatively prime to 5. One quickly computes that l > k , but
such elements generate a subgroup of order 5k , which is not large enough to
satisfy the condition of Theorem 1.1, Statement 3.
4
The Infinite Cyclic Cover of 74
The goal of this section is to prove the following result. It, along with Theorem
3.1, shows that 74 cannot be changed into a slice knot with a single crossing
change.
Theorem 4.1 If a crossing change converts 74 into a slice knot J , then the
corresponding curve S1 in M (74 ) is null homologous in H1 (L(15, 4), Z).
Before beginning the proof we need to set up notation and prove a lemma.
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Charles Livingston
The infinite cyclic cover of J is built from the infinite cyclic cover of the unknot,
U , by performing equivariant surgery on three families of curves: {K̃i }, {K̃i0 }
and {S̃i }, using the notation as before. (In each case, i = −∞, . . . , ∞.)
Following Rolfsen [R], one can draw that cover with the {K̃i }, {K̃i0 } drawn
explicitly, and the {S̃i } unknown curves. From this one finds the presentation
matrix of the infinite cyclic cover of J as a Z[t, t−1 ] module, with respect to
the basis given by the meridians of K̃0 , K̃00 and S̃0 , say k0 , k00 , and s0 . The
resulting presentation is given by the matrix


−2t + 3 − 2t−1 1 g(t)
A=
1
−2
0 .
−1
g(t )
0 f (t)
Here g(t) is an unknown polynomial describing the linking between the lifts of
S and those of K . (Notice that the lifts of S do not link the lifts of K 0 , since
τ (and so S ) misses the small disk bounded by K 0 and this disk lifts to a series
of disjoint disks bounded by the K̃i0 in the infinite cyclic cover.) Also, f (t) is
an unknown symmetric polynomial describing the self–linking of the lifts of S .
(It might be helpful for the reader to note that if g = 0 and f = 1 then the
determinant of the matrix is 4t − 7 + 4t−1 , the Alexander polynomial of 74 .)
Although g and f are unknown, two observations are possible. The first is that
f (1) = ±1; this is because ±1 surgery is being performed on S . The second
is that g(1) = 0, or that (t − 1) divides g , which follows from the fact that S
and K have 0 linking number, since S bounds the disk D in the complement
of K .
Lemma 4.2 If J is slice, then 4t − 7 + 4t−1 = ∆74 divides g .
Proof The determinant of A is given by
∆J (t) = f (t)∆74 (t) + 2g(t)g(t−1 ).
Since J is assumed to be slice we can rewrite this as
±H(t)H(t−1 ) = f (t)∆74 (t) + 2g(t)g(t−1 )
for some H(t). Clearly, if H is divisible by ∆74 then g(t) would also be and
we would be done. So, assume that neither H or g has factor ∆74 .
Working modulo ∆74 we now have the equation:
(∗)
2g(t)g(t−1 ) = ±H(t)H(t−1 ) ∈ Z[t, t−1 ]/h4t − 7 + 4t−1 i.
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The slicing number of a knot
√
There√is an injection φ : Z[t, t−1 ]/h4t − 7 + 4t−1 i → Q( −15) with φ(t±1 ) =
(7 ± −15)/4.
It follows
√
√ that if equation (∗) holds then we could factor 2 =
±( ac + bc −15)(( ac − bc −15) with a, b, and c integers with gcd(a, b, c) = 1.
Simplifying we would have
±2c2 − a2 − 15b2 = 0.
Working modulo 5 and using that ±2 is not a quadratic residue modulo 5, one
sees immediately that a and c are both divisible by 5, which implies (working
modulo 25) that b is divisible by 5 as well. Write a = 5s a0 , b = 5t b0 and
c = 5r c0 , with a0 , b0 , and c0 relatively prime to 5. Hence:
±2(52s c0 ) − 52t a0 − 3(52r+1 b0 ) = 0.
2
2
2
If among the three exponents of 5 that appear in this equation there is a unique
smallest exponent, then factoring out that power of 5 leaves an equation that
clearly cannot hold modulo 5. Hence, there must be two exponents that are
equal, and these must be the two even exponents. Factoring these out leaves
the equation:
0
2
2
2
±2c0 − a0 − 3(52r +1 b0 ) = 0.
Again using that ±2 is not a quadratic residue modulo 5 gives a contradiction.
We can now prove Theorem 4.1.
Proof of Theorem 4.1 The polynomial g determines the linking numbers of
the lifts of K and S to the n-fold cyclic branched cover of S 3 branched over
U as follows. Call the lifts K̄i and S̄i with i running from 0 to n − 1. The
linking numbers are given by equivariance and
lk(K̄0 , S̄i ) = ḡi
where ḡi is the coefficient of ti in the reduction ḡ of g to Z[t, t−1 ]/htn − 1i.
In the case of the 2–fold cover we are hence interested in P
the even and odd
index coefficients. For any integral polynomial F (x) =
ai ti the sum of
the even index coefficients is given by (F (1) + F (−1))/2 and the sum of the
odd index coefficients is (F (1) − F (−1))/2. In our case we have seen that
g(t) = (t − 1)(4t2 − 7t + 4)h(t) for some h. Hence, the sum of the even (or odd)
coefficients is given by ±15h(−1). In particular, the linking number is divisible
by 15. Hence S̄i = Si is null homologous in the L(15, 4) obtained by surgery
on K1 .
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Charles Livingston
Extensions
The proof that 74 has slicing number 2 clearly generalizes to other knots, though
a general statement is somewhat technical. On the other hand, these methods
seem not to apply effectively in addressing the next level of complexity—finding
a knot K with gs (K) = 2 but with slicing number 3.
Conjecture 5.1 The difference us (K) − gs (K) can be arbitrarily large.
In fact, this gap should be arbitrarily large even for knots with gs = 1.
In retrospect, Askitas’s question was optimistic. It is easily seen that if a
knot can be converted into a slice knot by making n positive and n negative
crossing changes, then gs (K) ≤ n. More generally, we have the following signed
unknotting number.
Definition 5.2 For a knot K , let I denote the set of pairs of nonnegative
integers (m, n) such that some collection of m positive crossing changes and
n negative crossing changes converts K into a slice knot. Define the invariant
Us (K) by
Us (K) = min {max(m, n)}
(m,n)∈I
The following result has an elementary proof.
Theorem 5.3 For all K , gs (K) ≤ Us (K).
The only bounds that we know of relating to Us are those arising from gs ,
and so it is possible that Us (K) = gs (K) for all K . However, a more likely
conjecture is the following.
Conjecture 5.4 The difference Us (K) − gs (K) can be arbitrarily large.
Even the following example is unknown.
Question Does Us (74 ) = 1?
The example we describe below indicates that proving that Us (74 ) = 2 may be
quite difficult.
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The slicing number of a knot
General Twisting One can think of performing a crossing change as grabbing
two parallel strands of a knot with opposite orientation and given them one full
twist. More generally, one can grab 2k parallel strands of K with k of the
strands oriented in each direction and giving them one full twist. Call this a
generalized crossing change. With a little care, the proof that 74 cannot be
converted into a slice knot generalizes to show the following:
Theorem 5.5 The knot 74 cannot be converted into a slice knot using a single
generalized crossing change.
On the other hand, consider Figure 2. The illustrated knot is slice since the
dotted curve on the Seifert surface is unknotted and has framing 0. If a right–
handed twist is put on the strands going through the circle labelled −1 and a
left–handed twist is put on the strands going through the circle labelled +1,
then the knot 74 results. Hence, 74 can be converted into a slice knot by
performing one positive and one negative generalized crossing change.
-1
+1
Figure 2: Twisting 74 to a slice knot
Since all the relevant techniques that we know of do not distinguish between
crossing changes and generalized crossing changes, the difficulty associated to
disproving showing that Us (74 ) = 2 is now clear.
It is worth pointing out here that clearly 74 can be converted into a slice knot
(actually the unknot) using two negative crossing changes, but an analysis of
signatures and a minor generalization of the results of [CL] shows that it cannot
be converted into a slice knot (or a knot with signature 0) using two positive
generalized crossing changes.
Related to this discussion we have the follow result. Its proof is a bit technical
to include here and will be described in detail elsewhere.
Theorem 5.6 A knot K with 3–sphere genus g(K) can be converted into the
unknot using 2g(K) generalized crossing changes.
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Charles Livingston
Addendum (December 15, 2002) It has been pointed out to the author
that results of Murakami and Yasuhara (Four-genus and four-dimensional clasp
number of a knot, Proc. Amer. Math. Soc. 128 (2000), no. 12, 3693–3699)
imply that gs (816 ) = 1 but us (816 ) = 2. The methods used there are different
from those of this paper.
References
[A]
N. Askitas. Multi-# unknotting operations: a new family of local moves on a
knot diagram and related invariants of knots, J. Knot Theory Ramifications 7
(1998), no. 7, 857–871.
[CG]
A. Casson and C. Gordon. Cobordism of classical knots. A la recherche de la
Topologie perdue, ed. by Guillou and Marin, Progress in Mathematics, Volume
62, 1986. (Originally published as Orsay Preprint, 1975.)
[CL]
T. Cochran and W. B. R. Lickorish. Unknotting information from 4-manifolds,
Trans. Amer. Math. Soc. 297 (1986), no. 1, 125–142
[G]
C. McA. Gordon. Some aspects of classical knot theory. Knot theory (Proc.
Sem., Plans-sur-Bex, 1977), pp. 1–60 Lecture Notes in Math., 685, Springer,
Berlin, 1978.
[K]
A. Kawauchi. Distance between links by zero-linking twists, Kobe J. Math. 13
(1996), no. 2, 183–190.
[L]
W. B. R. Lickorish. The unknotting number of a classical knot, Combinatorial
methods in topology and algebraic geometry (Rochester, N.Y., 1982), 117–121,
Contemp. Math., 44, Amer. Math. Soc., Providence, RI, 1985.
[Mo]
J. Montesinos. Surgery on links and double branched covers of S 3 , Knots,
groups, and 3-manifolds (Papers dedicated to the memory of R. H. Fox), 227–
259. Ann. of Math. Studies, No. 84, Princeton Univ. Press, Princeton, N.J.,
1975.
[Mu]
K. Murasugi. On a certain numerical invariant of link types, Trans. Amer. Math.
Soc. 117 (1965) 387–422.
[N]
Y. Nakanishi. A note on unknotting number, Math. Sem. Notes Kobe Univ. 9
(1981), no. 1, 99–108.
[O]
T. Ohtsuki. Problems on Invariants of Knots and 3–Manifolds, Invariants of
Knots and 3–Manifolds, Kyoto University 2001. Geometry and Topology Monographs, Volume 4 (2002).
[R]
D. Rolfsen. Knots and Links. Publish or Perish, Berkeley CA (1976).
Department of Mathematics, Indiana University, Bloomington, IN 47405, USA
Email: livingst@indiana.edu
Algebraic & Geometric Topology, Volume 2 (2002)
Received: 13 June 2002